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Since $\mathbf E = -∇Φ - ∂\mathbf A/∂t$ one expects an oscillating $\mathbf E$ field even in the null of a Hertzian Dipole unless the two right hand side terms cancel -- which they do in the far field of the null.

However, in the near field of the null, the terms do not completely cancel, leaving a residual oscillating E-field.

Since the null has, by definition, no $∇ × \mathbf A$ curl in the oscillating $\mathbf A$, there is no $\mathbf B$ thence no $\mathbf H$ field and therefore no $\mathbf E × \mathbf H$ and since $\mathbf E × \mathbf H$ is the only accepted definition for the dipole's Poynting vector, there is no accepted way for energy to be locally available at points along the dipole's null.

If one places a particle of charge $q$ and mass $m$ along the null, it must experience a force, $\mathbf F=q\mathbf E$ and thence acceleration $\mathbf F=m\mathbf a$.

Where does this energy come from, and how is it delivered without violating locality?

Community
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James Bowery
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    You don't need to guess what the field of the dipole is from qualitative considerations on the potential. If a point dipole oscillates at frequency $\omega=ck$, then the (complex) electric field it produces is $$ \mathbf E=\frac{1}{4\pi\varepsilon_0}\left[ k^2(\hat{\mathbf r}\times \mathbf p)\times\hat{\mathbf r}\frac{1}r+(3\hat{\mathbf r}(\hat{\mathbf r}·\mathbf p)-\mathbf p)\left(\frac1{r^3}-\frac{ik}{r^2}\right) \right]e^{ikr-\omega t} $$ (from Jackson, 3rd ed., eq. 9.18). There is indeed a nonzero near-field component along the direction of $\mathbf p$. – Emilio Pisanty Sep 16 '15 at 20:00
  • If the oscillation is not monochromatic then this changes, but I don't think that's crucial to your (very interesting) question. – Emilio Pisanty Sep 16 '15 at 20:00
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    You can find a much more simple field configuration with the same problem: a static electric field. It, too, will accelerate charges, but there is no energy flow caused by the electric field itself. From that I would conclude that the Poynting vector of the field without test charge is not needed, at all, to extract energy from a static field. It may be an interesting question to know if the Poynting vector of the field with test charge is... it does, indeed, sound like a non-trivial self-consistency problem of electrodynamics, similar to that of the self-energy of charges. – CuriousOne Sep 16 '15 at 20:09
  • Isn't there still a local energy density: $E^2$? – Shane P Kelly Sep 16 '15 at 20:21

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The example by CuriousOne is spot on - we know charged particle will accelerate in static external electric field, so generally, Poynting vector based on the external field does not need to be non-zero when charged particle gain kinetic energy.

The apparent problem with local energy conservation is caused by using the wrong expression for the energy flux density - you seem to assume it is a function of external fields only. The exact form of the expression depends on whether the charged particles are points or extended bodies, but in both cases it is not a function of external fields only.

For example, if the charged particles are extended (charge and current density are finite), the Poynting theorem is valid. The local version of this theorem states that

$$ \mathbf j\cdot\mathbf E = -\frac{\partial}{\partial t}\left(\frac{1}{2}\epsilon_0 E^2 + \frac{1}{2\mu_0} B^2\right) - \nabla \cdot (\mathbf E\times\mathbf B/\mu_0) $$

where $\mathbf j$ is electric current density and $\mathbf E$,$\mathbf B$ are total EM fields. One cannot use only external electric and magnetic field and expect this equation remains valid.

When a charged particle starts accelerating in electrostatic external field, external magnetic field indeed is zero, but total magnetic field is not. This is because accelerated particle itself has non-zero magnetic field in its vicinity.

For example, if fields of the particle are assumed retarded, the region of space where the magnetic field is non-zero will be a sphere centered at the position of the particle when it started accelerating and its boundary will expand with speed of light.

Ján Lalinský
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I believe this apparent contradiction to stem from a misunderstanding that energy can be transferred only by the one mechanism to which the Poynting vector applies:

The Poynting vector is defined as ExH, and applies to a "launched" electromagnetic wave in propagation. In your example, the energy is being transferred by a quasi-static E-field in the absence of a B-field. There is no H, therefore there is no ExH, hence there is no Poynting vector. The Poynting vector does not apply to your example. N.B. the lack of an electromagnetic wave described by a Poynting vector in no way implies that energy cannot be transferred via other means. Energy may be transferred in many ways.

E.g. kinetic energy transferred to a test-charge via a static electric field is always transferred via means other than those described by the Poynting vector. Ditto transfer of energy via lone magnetic, weak, strong, or gravitational fields, or mechanical, acoustic, thermodynamic, etc. mechanisms.

godot
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  • As Emilio Pisanty points out above, there are 3 exponents of r: 1/r^1, 1/r^2 and 1/r^3. "Quasi-static" is associated with 1/r^3. Is it also associated with 1/r^2, which is important in the far field "cancellation" of $\mathbf 0 = -∇Φ - ∂\mathbf A/∂t$ ? Perhaps even, in some sense not usually considered, with 1/r^1? – James Bowery Sep 28 '15 at 14:57
  • From "Electromagnetics from a quasistatic perspective" http://www.if.ufrj.br/~dore/Fis3/Electromagnetics_from_a_quasistatic_perspective.pdf "An important feature of quasistatics is instantaneous interaction at a distance."

    Invoking "quasistatics" to explain things where the test charge is further out than several wavelengths but not yet in the "far field" begs the question about "locality". Indeed, CuriousOne opines that even in the 1/r^3 regime, there appears to be a genuine self-consistency problem.

     – James Bowery Sep 28 '15 at 15:03